Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $p \neq 0$. $k = \dfrac{p^3 + 9p^2 + 18p}{4p^3 + 28p^2 + 24p} \times \dfrac{-2p + 2}{p + 3} $
Answer: First factor out any common factors. $k = \dfrac{p(p^2 + 9p + 18)}{4p(p^2 + 7p + 6)} \times \dfrac{-2(p - 1)}{p + 3} $ Then factor the quadratic expressions. $k = \dfrac {p(p + 6)(p + 3)} {4p(p + 6)(p + 1)} \times \dfrac {-2(p - 1)} {p + 3} $ Then multiply the two numerators and multiply the two denominators. $k = \dfrac { p(p + 6)(p + 3) \times -2(p - 1)} { 4p(p + 6)(p + 1) \times (p + 3)} $ $k = \dfrac {-2p(p + 6)(p + 3)(p - 1)} {4p(p + 6)(p + 1)(p + 3)} $ Notice that $(p + 6)$ and $(p + 3)$ appear in both the numerator and denominator so we can cancel them. $k = \dfrac {-2p\cancel{(p + 6)}(p + 3)(p - 1)} {4p\cancel{(p + 6)}(p + 1)(p + 3)} $ We are dividing by $p + 6$ , so $p + 6 \neq 0$ Therefore, $p \neq -6$ $k = \dfrac {-2p\cancel{(p + 6)}\cancel{(p + 3)}(p - 1)} {4p\cancel{(p + 6)}(p + 1)\cancel{(p + 3)}} $ We are dividing by $p + 3$ , so $p + 3 \neq 0$ Therefore, $p \neq -3$ $k = \dfrac {-2p(p - 1)} {4p(p + 1)} $ $ k = \dfrac{-(p - 1)}{2(p + 1)}; p \neq -6; p \neq -3 $